What is the integral from 0 to pi/2 of (sin^2(θ))/2 ?

The integral from 0 to pi/2 of (sin^2(θ))/2 is pi/8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to pi/2 of (sin^2(θ))/2

\int_{0}^{\frac{π}{2}} \frac{sin ^{2} ( θ )}{2} d θ

\frac{π}{8}

Decimal Notation

0.39269 \dots

Solution steps

\int_{0}^{\frac{π}{2}} \frac{sin ^{2} ( θ )}{2} d θ

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{0}^{\frac{π}{2}} sin^{2} (θ) d θ

Rewrite using trig identities

= \frac{1}{2} \cdot \int_{0}^{\frac{π}{2}} \frac{1 - cos ( 2 θ )}{2} d θ

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int_{0}^{\frac{π}{2}} 1 - cos (2 θ) d θ

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} \cdot \frac{1}{2} (\int_{0}^{\frac{π}{2}} 1 d θ - \int_{0}^{\frac{π}{2}} cos (2 θ) d θ)

\int_{0}^{\frac{π}{2}} 1 d θ = \frac{π}{2}

\int_{0}^{\frac{π}{2}} cos (2 θ) d θ = 0

= \frac{1}{2} \cdot \frac{1}{2} (\frac{π}{2} - 0)

Simplify

\frac{1}{2} \cdot \frac{1}{2} (\frac{π}{2} - 0) : \frac{π}{8}

= \frac{π}{8}

\int_{x^{2}}^{x^{6}} (2 t - 1)^{3} d t

\int_{0}^{4} (3 t - 8) d t

\int_{2}^{6} x^{2} - 3 x d x

\int_{- 2}^{1} (3 - x^{2}) - (x + 1) d x

\int_{0}^{t} x e^{- x^{2}} d x

What is the integral from 0 to pi/2 of (sin^2(θ))/2 ?
The integral from 0 to pi/2 of (sin^2(θ))/2 is pi/8